Let $K$ be a field, $R=K[x^2,x^3]$, $S=K[x]$, and consider $S$ as an $R$-module. Given $f: S \to R \oplus R$ so that $f:p \mapsto (x^3p,-x^2p)$, prove that $f\otimes 1: S \otimes_R S \to (R\oplus R) \otimes_R S$ is not injective.
Asked
Active
Viewed 166 times
1 Answers
4
You actually have a short exact sequence $0\to S\stackrel{f}\to R^2\stackrel{g}\to S\to 0$, where $g(a,b)=X^2a+X^3b$, and by tensoring this with $S$ want to prove that it is not exact, that is, $S$ is not $R$-flat.
Since $R^2\otimes_RS\simeq S^2$ by $(a,b)\otimes c\mapsto(ac,bc)$, we can see $f\otimes 1: S \otimes_R S \to S^2$ sending $p\otimes_R q$ to $(X^3pq,-X^2pq)$. Now notice that $(f\otimes 1)(p\otimes q)=(f\otimes 1)(1\otimes pq)$ for any $p,q\in S$. Set $p=X$ and $q=X+1$. Our job is to show that $X\otimes (X+1)\ne 1\otimes X(X+1)$ in $S \otimes_R S$. Consider the surjective homomorphism $S \otimes_R S \to S \otimes_R S/XS$ which sends $X\otimes (X+1)$ to $X\otimes 1$ and $1\otimes X(X+1)$ to $0$. Now we have to prove that $X\otimes 1\ne 0$ in $S \otimes_R S/XS$. We have $R\to S\to S/XS$ a surjective homomorphism whose kernel is $(X^2,X^3)R$, so $S/XS\simeq R/(X^2,X^3)R$, so $S \otimes_R S/XS\simeq S \otimes_R R/(X^2,X^3)R\simeq S/(X^2,X^3)S=S/X^2S$. Among so many isomorphisms we still can see that $X\otimes 1$ corresponds to the residue class of $X$ in $S/X^2S$, and thus it is not zero.
user26857
- 52,094
-
I forgot to say that there is a simpler way to show that $S$ is not $R$-flat: $(X^2R\cap X^3R)S\ne X^2S\cap X^3S$. – user26857 Dec 07 '14 at 21:21
-
I'm sorry, I don't get why $(f\otimes 1)(p\otimes_Rq)=(f\otimes 1)(1\otimes_Rpq)$. – Leo163 Dec 07 '14 at 23:20
-
1@Leo163 By definition $(f\otimes 1)(u\otimes_R v)=(X^3uv,-X^2uv)$. – user26857 Dec 07 '14 at 23:43